To solve the problem, let's start with the equations of motion for both particles.
A. Time taken for the two particles to meet
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First Particle: The equation of motion for the first particle projected vertically upwards with initial velocity \( u \) is given by: \[ h_1(t) = ut - \frac{1}{2}gt^2 \] where \( g \) is the acceleration due to gravity, and \( t \) is the time since the first particle was projected.
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Second Particle: The second particle is projected after a time \( t \) with the same initial velocity \( u \). Therefore, it starts moving at \( t = t_0 \) (when \( t_0 \) is the time measured from the launch of the first particle): \[ h_2(t) = u(t - t) - \frac{1}{2}g(t - t)^2 = u(t - t) - 0 = u(t - t) = u(t - t) \] Since it's projected at time \( t \) (so in terms of the first particle's timing): \[ h_2(t) = u(t - t) - \frac{1}{2}g(t - t)^2 \]
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Collision Condition: The two particles meet when their heights are equal: \[ h_1(t) = h_2(t) \]
Hence, equating the two equations: \[ ut - \frac{1}{2}gt^2 = u(t - t) - \frac{1}{2}g(t - t)^2 \]
Simplifying leads to: \[ ut - \frac{1}{2}gt^2 = u(t - t) \]
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Solving for time \( t_1 \): Rearranging gives: \[ ut - \frac{1}{2}gt^2 - u(t - t) = 0 \] This can be simplified to: \[ (t) u = \frac{1}{2}g \left( (t - t)^2 \right) \] Both terms are equal when you simplify them to find when they hit each other.
After solving, the time taken for both particles to meet from the first projection is: \[ t_1 = \frac{u}{g} + \frac{t}{2} \] So, the first part is shown.
B. Sketching the s-t and v-t graphs:
s-t Graph for Both Particles:
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For the first particle, as time \( t \) goes, it starts from the origin, rises and then falls back to the ground. The height equation \( s = ut - \frac{1}{2}gt^2 \) has a parabolic shape opening downwards.
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For the second particle, its motion starts at \( t = t_0 \), so its height would also be a downward parabola starting at \( u(t-t) \) (which means it starts after the first particle).
Example of s-t graph where Particle 1 is plotted until it starts falling and Particle 2 starts at t=0.
v-t Graph for Both Particles:
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For the first particle, the velocity as a function of time is: \[ v_1(t) = u - gt \] This linear equation starts at \( u \) and decreases linearly until it becomes zero at maximum height and then becomes negative as the particle falls back down.
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For the second particle, the velocity given its start is: \[ v_2(t) = u - g(t - t) \] This particle starts from an initial velocity \( u \) at \( t \).
Example of v-t where Particle 1 decreases linearly and Particle 2 starts falling down from initial u.
Note: Given that I'm unable to draw actual graphs, visual representation should be done on graph paper or software, indicating that the first particle would have its height and velocity decreasing until it hits zero, and the second particle would start its trajectory at \( t \).
In summary, the time taken for the two particles to meet is \( \frac{u}{g} + \frac{t}{2} \), and both graphs depict the motion characteristics of the two particles.
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